What is (x+4)^2+((y-3)/4)^2=1 ?

The solution to (x+4)^2+((y-3)/4)^2=1 is Ellipse with (h,k)=(-4,3),a=1,b=4

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(x+4)^2+((y-3)/4)^2=1

(x + 4)^{2} + (\frac{y - 3}{4})^{2} = 1

(h, k) = (- 4, 3), a = 1, b = 4

Solution steps

(x + 4)^{2} + (\frac{y - 3}{4})^{2} = 1

Rewrite

(x + 4)^{2} + (\frac{y - 3}{4})^{2} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 4 ) ) ^{2}}{1 ^{2}} + \frac{( y - 3 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 4, 3), a = 1, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 4, 3), b = 4, a = 1

\frac{( x - 1 ) ^{2}}{5} + \frac{( y - 3 ) ^{2}}{4} = 1

9 x^{2} + 4 y^{2} - 126 x - 40 y + 505 = 0

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{25} = 1

\frac{( x + 1 ) ^{2}}{1} + \frac{y ^{2}}{4} = 1

\frac{y ^{2}}{4 - x ^{2}} = 1

What is (x+4)^2+((y-3)/4)^2=1 ?
The solution to (x+4)^2+((y-3)/4)^2=1 is Ellipse with (h,k)=(-4,3),a=1,b=4